🔙 Prerequisite Revision:
👉 Chapter 2: Electrostatic Potential & CapacitanceChapter 3: Current Electricity
Vol 1: Electric Current & Ohm's Law
Electric Current ($I$): The rate of flow of electric charge. $I = q/t$.
Unit: Ampere (A). Scalar Quantity.
Drift Velocity ($v_d$)
The average velocity with which free electrons get drifted towards the positive end of the conductor under the influence of an external electric field.
($\tau$ = Relaxation Time)
Relation between Current and Drift Velocity:
$$I = n e A v_d$$
Ohm's Law
At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.
Vol 2: Resistivity & Temperature Dependence
Resistance ($R$): Opposition to current flow. $R = \rho \frac{l}{A}$.
Resistivity ($\rho$): Intrinsic property of material. Independent of dimensions.
$$\vec{J} = \sigma \vec{E}$$
(Where $J$ = Current Density, $\sigma$ = Conductivity = $1/\rho$)
Temperature Dependence
Resistance increases with temperature for conductors.
($\alpha$ = Temperature Coefficient of Resistance)
Vol 3: Cells & Kirchhoff's Rules
EMF ($\varepsilon$): Potential difference when circuit is open.
Terminal Voltage ($V$): Potential difference when circuit is closed.
Relation: $V = \varepsilon - Ir$ (During discharge).
Kirchhoff's Rules (KCL & KVL)
$\sum I_{in} = \sum I_{out}$
(Conservation of Charge)
$\sum \Delta V = 0$ in a closed loop.
(Conservation of Energy)
Vol 4: Wheatstone Bridge
An arrangement of 4 resistors used to determine unknown resistance.
[attachment_0](attachment)When no current flows through Galvanometer ($I_g = 0$):
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$
Note: The Slide Wire Bridge (Meter Bridge) works on this principle.
Vol 5: Mission 100 Question Bank
Section A: MCQs
Q1. Kirchhoff's Loop Rule is a consequence of conservation of:
(a) Charge (b) Momentum (c) Energy (d) Mass
Ans: (c) Energy.
Q2. If the temperature of a conductor increases, its drift velocity:
(a) Increases (b) Decreases (c) Remains same
Ans: (b) Decreases (Relaxation time $\tau$ decreases due to more collisions).
Section B: Numericals (Important)
Q3. A battery of EMF 10 V and internal resistance $3 \Omega$ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage?
Solution:
Given: $\varepsilon = 10$ V, $r = 3 \Omega$, $I = 0.5$ A.
1. $\varepsilon = I(R+r) \Rightarrow 10 = 0.5(R + 3)$
$20 = R + 3 \Rightarrow R = 17 \Omega$.
2. $V = \varepsilon - Ir = 10 - (0.5 \times 3) = 10 - 1.5 = 8.5$ V.
Ans: $17 \Omega$, 8.5 V.
Section C: Conceptual
Q4. Why is a potentiometer preferred over a voltmeter for measuring EMF?
Ans: A potentiometer works on the null deflection method (draws no current from the cell at balance point), so it measures the true EMF. A voltmeter always draws some current, measuring terminal voltage ($V < \varepsilon$).
🎉 SERIES COMPLETED! 🎉
Chapter 1 to 14: All Chapters are now Ready.
Your Physics "Mahagranth" is fully interlinked and complete.


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